Question

Two point charges, 3q and -q, are separated by a distance a. For each of the arrangements in Fig. 3.35, find (i) the monopole moment, (ii) the dipole moment, and (iii) the approximate potential (in spherical coordinates) at large r (include both the monopole and dipole contributions).

Short answer

Answer

1. The total potential at a distance r including monopole term is $\frac{\mathbf{4}\mathbf{\theta }}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\left[\frac{2q}{r}+\frac{3qacos}{r^2}\right]$.

2. The total potential at a distance r including monopole term is $\frac{\mathbf{4}\mathbf{\theta }}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\left[\frac{2q}{r}+\frac{qacos}{r^2}\right]$.

3. The total potential at a distance r including monopole term is $\frac{\mathbf{4}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\left[\frac{2q}{r}+\frac{3qasin\varphi }{r^2}\right]$.

Step-by-step solution

Given data

a)

From the above figure,

Monopole moment

$Q=3q=2q$

Dipole moment

$$\begin{gathered} p=\left[3qaz+\left(-q\right)0\right] \\ p=3qaz \end{gathered}$$

Determine monopole moment

Write the expression for total potential at a distanceincluding monopole term.

$$\begin{gathered} V\left(r\right)=V_{mQnQ}+Vdipole \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{Q}{r}+\frac{1}{4\pi {\epsilon }_0}\frac{p\widehat{·}r}{r^2} \\ =\frac{1}{4\pi {\epsilon }_0}\left[\frac{2q}{r}+\frac{3qa\cos \theta }{r^2}\right] \end{gathered}$$

Here, $p\widehat{·}r=3qa\cos \theta$

Therefore, the total potential at a distance r including monopole term is $\frac{1}{4\pi {\epsilon }_0}\left[\frac{2q}{r}+\frac{3qa\cos \theta }{r^2}\right]$.

Determine dipole moment

b)

From the above figure,

Monopole moment

$Q=2q$

Dipole moment

$$\begin{gathered} p=-qa\left(\widehat{-}z\right) \\ =qaz \end{gathered}$$

Write the expression for total potential at a distance r including monopole term.

$$\begin{gathered} V\left(r\right)=V_{mono}+Vdipole \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{Q}{r}+\frac{1}{4\pi {\epsilon }_0}\frac{\rho \widehat{·}r}{r^2} \end{gathered}$$

Substitute $2q$for Q and $3qa\cos \theta$for $p\widehat{·}r$.

$V\left(r\right)=\frac{1}{4\pi {\epsilon }_0}\left[\frac{2q}{r}+\frac{3qa\cos \theta }{r^2}\right]$

Therefore, the total potential at a distance $r$including monopole term is

$\frac{1}{4\pi {\epsilon }_0}\left[\frac{2q}{r}+\frac{3qa\cos \theta }{r^2}\right]$.

Determine potential

c)

From the above figure,

Monopole moment

$Q=2q$

Dipole moment

$p=3qa\hat{y}$

Write the expression for total potential at a distance r including monopole term.

$\begin{array}{rcl}V\left(r\right)& =& V_{mono}+V_{dipole}\\ & =& \frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}+\frac{1}{4\pi {\epsilon }_0}\frac{p·\hat{r}}{r^2}\\ & =& \frac{1}{4\pi {\epsilon }_0}\left[\frac{2q}{r}+\frac{3qa\sin \theta \sin \varphi }{r^2}\right]\\ & & \end{array}$

Therefore, the total potential at a distance r including monopole term is

$\frac{1}{4\pi {\epsilon }_0}\left[\frac{2q}{r}+\frac{3qa\sin \theta \sin \varphi }{r^2}\right]$

As,

$\hat{r}=\sin \theta \cos \varphi \hat{x}+\sin \theta \sin \varphi \hat{y}+\cos \theta \hat{z}$

Then,

$\hat{y}·\hat{r}=\sin \theta \sin \varphi$.